Question

A DISTRIBUIÇÃO DE IDADE DE UMA COMUNIDADE É FORNECIDA PELA TABELA: (MÉDIA, MEDIANA E MODA) AJUDA, POR FAVOR!!!

Answer 1

desculpa mas a única q eu consegui foi a

C) média:2,1

Mediana:10,12,13,"22",22,31,50 ( 22+22 =44_2 = (8)

Modo:22

Answer 2

Explicação passo-a-passo:

a) Variável quantitativa

b)

$\sf \Big|~~idade~~~~\Big|~~~~~freq.~~~~~\Big|~~~~freq.~ab.~~~\Big|~~~~~~freq.~~~~~~\Big|~~~~freq.~rel~~~~</p><p>\Big|$

$\sf \Big|~~(anos)~~\Big|~~absoluta~~\Big|~~acumulada~\Big|~~~~relativa~~~\Big|~~acumulada~~\Big|$

$\sf \Big|~~~~~~0~~~~~~\Big|~~~~~~~~12~~~~~~~\Big|~~~~~~~~~12~~~~~~~~\Big|~~~~~7,5\%~~~~~\Big|~~~~~~~7,5\%~~~~~~\Big|$

$\sf \Big|~~~~~~1~~~~~~\Big|~~~~~~~~13~~~~~~~\Big|~~~~~~~~~25~~~~~~~~\Big|~~~8,125\%~~~\Big|~~~~15,625\%~~~\Big|$

$\sf \Big|~~~~~~2~~~~~~\Big|~~~~~~~~22~~~~~~~\Big|~~~~~~~~~47~~~~~~~~\Big|~~~13,75\%~~~\Big|~~~29,375\%~~~~\Big|$

$\sf \Big|~~~~~~3~~~~~~\Big|~~~~~~~~50~~~~~~~\Big|~~~~~~~~~97~~~~~~~~\Big|~~~31,25\%~~~\Big|~~~60,625\%~~~~\Big|$

$\sf \Big|~~~~~~4~~~~~~\Big|~~~~~~~~31~~~~~~~\Big|~~~~~~~~128~~~~~~~\Big|~~19,375\%~~\Big|~~~~~~~80\%~~~~~~~\Big|$

$\sf \Big|~~~~~~5~~~~~~\Big|~~~~~~~~22~~~~~~~\Big|~~~~~~~~150~~~~~~~\Big|~~~13,75\%~~~\Big|~~~~93,75\%~~~~\Big|$

$\sf \Big|~~~~~~6~~~~~~\Big|~~~~~~~~10~~~~~~~\Big|~~~~~~~~160~~~~~~~\Big|~~~~6,25\%~~~~\Big|~~~~~~100\%~~~~~~\Big|$

c)

=> Média

$\sf M=\dfrac{0\cdot12+1\cdot13+2\cdot22+3\cdot50+4\cdot31+5\cdot22+6\cdot10}{12+13+22+50+31+22+10}$

$\sf M=\dfrac{0+13+44+150+124+110+60}{160}$

$\sf M=\dfrac{501}{160}$

$\sf \red{M=3,13125}$

=> Mediana

A mediana é valor central (ou a média dos valores centrais) quando colocamos os valores em ordem crescente

Como há 160 valores, os valores centrais são o 80° e o 81°

Temos $\sf 12+13+22=47$. Assim, 47 crianças tem menos de 3 anos.

E $\sf 12+13+22+50=97$. Note que já passamos de 81. Os valores centrais são 3 e 3

Logo, a mediana é $\sf \dfrac{3+3}{2}=\dfrac{6}{2}=3$

A mediana é 3

=> Moda

É o valor que mais se repete.

A moda é 3

d)

=> Variância

$\sf \sigma^2=\dfrac{12\cdot(0-3,13125)^2+13\cdot(1-3,13125)^2+22\cdot(2-3,13125)^2+50\cdot(3-3,13125)^2+31\cdot(4-3,13125)^2+22\cdot(5-3,13125)^2+10\cdot(6-3,13125)^2}{12+13+22+50+31+22+10}$

$\sf \sigma^2=\dfrac{12\cdot(-3,13125)^2+13\cdot(-2,13125)^2+22\cdot(-1,13125)^2+50\cdot(0,13125)^2+31\cdot(-1,13125)^2+22\cdot2,13125^2+10\cdot3,13125^2}{160}$

$\sf \sigma^2=\dfrac{12\cdot9,8047+13\cdot4,5422+22\cdot1,2797+50\cdot0,0172+31\cdot1,2797+22\cdot4,5422+10\cdot9,8047}{160}$

$\sf \sigma^2=\dfrac{117,6564+59,0486+28,1534+0,86+39,6707+99,9284+98,047}{160}$

$\sf \sigma^2=\dfrac{443,3645}{160}$

$\sf \red{\sigma^2=2,771}$

=> Desvio padrão

$\sf \sigma=\sqrt{\sigma^2}$

$\sf \sigma=\sqrt{2,771}$

$\sf \red{\sigma=1,6646}$

=> Coeficiente de variação

$\sf CV\%=\dfrac{\sigma}{M}\cdot100$

$\sf CV\%=\dfrac{1,6646}{3,13125}\cdot100$

$\sf CV\%=\dfrac{166,46}{3,13125}$

$\sf CV\%=\dfrac{16646000}{313125}$

$\sf \red{CV\%=53,16\%}$

Alta dispersão, dados heterogêneos